h2 / a level mathematics november 2007 paper 2 (9740/2) · h2/ "a" level mathematics...

H2 / ”A” Level MathematicsNovember 2007 Paper 2 (9740/2)

Kelvin Sohhttp://alevelmathblog.wordpress.com/

[email protected]

June 22, 2013

Contents

Question 1 2

Question 2 2

Question 3 4

Question 4 6

Question 5 7

Question 6 8

Question 7 9

Question 8 10

Question 9 11

Question 10 11

Question 11 12

1

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H2/ ”A” Level Mathematics November 2007 Paper 2 (9740/2).http://alevelmathblog.wordpress.com/ 2

Question 1Let x, y and z be the price per kg of pineapples, mangoes and lycheesrespectively.

Suresh: 1.15x+ 0.60y + 0.55z = 8.28

Fandi: 1.20x+ 0.45y + 0.30z = 6.84

Cindy: 2.15x+ 0.90y + 0.65z = 13.05

From the G.C.,

x = 3.5, y = 2.6, z = 4.9

∴ Amount Lee Lian paid = $(3.5× 1.30 + 2.6× 0.25 + 4.9× 0.50)

= $7.65.

Question 2

(i) Let pn be the statement un = 1n2 for n ≥ 1, n ∈ Z.

p1 : LHS = u1 = 1

RHS =1

12= 1

LHS = RHS

Hence p1 is true.



Question 2 (cont.)

Assume that pk is true for some k ∈ Z+ (i.e. uk = 1k2

)

To prove: pk+1 : uk+1 =1

(k + 1)2

LHS = uk+1

= uk −2k + 1

k2(k + 1)2

=1

k2− 2k + 1

k2(k + 1)2(by assumption)

=k2 + 2k + 1− 2k − 1

k2(k + 1)2

=k2

k2(k + 1)2

=1

(k + 1)2

= RHS

Hence pk is true ⇒ pk+1 is true.p1 is also true.Hence, by mathematical induction, un = 1

n2 ∀n ∈ Z+.

(ii) N∑n=1

2n+ 1

n2(n+ 1)2=

N∑n=1

(un − un+1)

= u1 − u2+ u2 − u3+ · · ·+ uN − uN+1

=u1 − uN+1 (by the method of differences)

=1− 1

(N + 1)2



Question 2 (cont.)

(iii) As N →∞, 1(N+1)2

→ 0

Hence the series is convergent and∞∑n=1

2n+ 1

n2(n+ 1)2= 1.

(iv) N∑n=1

2n+ 1

n2(n+ 1)2= 1− 1

(N + 1)2

Replace n with n− 1N+1∑n=2

2(n− 1) + 1

(n− 1)2(n− 1 + 1)2= 1− 1

(N + 1)2

N+1∑n=2

2n− 1

n2(n− 1)2= 1− 1

(N + 1)2

Replace N with N − 1N∑n=2

2n− 1

n2(n− 1)2= 1− 1

(N)2

Question 3

(i) Let f(x) = (1 + x)n.f ′(x) = n(1 + x)n−1

f ′′(x) = n(n− 1)(1 + x)n−2

f ′′′(x) = n(n− 1)(n− 2)(1 + x)n−3

f(x) = f(0) +f ′(0)

1!x+

f ′′(0)

2!x2 +

f ′′′(0)

3!x3 + · · ·

= 1 + nx+n(n− 1)

2x2 +

n(n− 1)(n− 2)

6x3 + · · ·



Question 3 (cont.)

(ii) (4− x)32 (1 + 2x2)

32

= 432 (1− x

4)32 (1 +

3

2(2x2) + · · · )

= 8

(1 +

3

2(−x

4) +

(32)(1

2)

2(−x

4)2 +

(32)(1

2)(−1

2)

6(−x

4)3 + · · ·

)(1 +

3x2 + · · · )= 8(1− 3

8x+

3

128x2 +

1

1024x3 + · · · )(1 + 3x2 + · · · )

= 8(1− 3

8x+

3

128x2 +

1

1024x3 + 3x2 − 9

8x3 + · · · )

= 8− 3x+387

16x2 − 1151

128x3 + · · ·

(iii) For the expansion to be valid,

|x| < 4 and |2x2| < 1

−4 < x < 4 x2 <1

2

− 1√2< x <

1√2

Combining,

−√

2

2< x <

√2

2



Question 4

(i)∫ 5

3π

0

sin2 xdx =

∫ 53π

0

1− cos 2x

2dx

=[x

2− sin 2x

4

] 53π

0

=5π

6−

sin 10π3

4

=5π

6+

sin π3

4

=5π

6+

√3

8

∫ 53π

0

sin2 xdx =

∫ 53π

0

(1− cos2 x)dx

=5π

3−∫ 5

3π

0

cos2 xdx

∴∫ 5

3π

0

cos2 xdx =5π

3−∫ 5

3π

0

sin2 xdx

=5π

3− (

5π

6+

√3

8)

=5π

6−√

3

8

(ii)



Question 4 (cont.)

(a)Area of R =

∫ π2

0

x2 sinxdx

=[− x2 cosx

]π2

0+

∫ π2

0

2x cosxdx

= 0 +[2x sinx

]π2

0−∫ π

2

0

2 sinxdx

= 2(π

2) sin

π

2− 0 +

[2 cosx

]π2

0

= π + 2(0− 1)

= π − 2 units2

(b)Volume required = π

∫ π2

0

(x2 sinx)2dx

= 5.391 (3 d.p.) units3 (from the G.C.)

Question 5

(i) A restaurant may be interested in what its patrons think of itsfood quality. Quota sampling can be used by interviewing thefirst 20 customers below 40 years old and the first 20 customersabove or exactly 40 years old who are willing to be interviewed.Quota sampling would be appropriate as it is easy to conduct andinformation can be obtained very quickly.

Quota sampling is a non-random sampling method. Hence, oneof its disadvantage is that the sample may not be indicative of thewhole population. For example, the interviewer may only interviewthe more friendly customers who may give higher ratings to therestaurant’s food compared to the average patron.

(ii) The actual representation of the different strata (customers aboveor exactly, and below 40 years old in our example) in our population(i.e. all patrons of the restaurant) may not be known. In thatscenario, it will be impossible to use stratified sampling.

It will be possible to use stratified sampling only if the actual rep-resentation of the different strata is known.



Question 6

Let X be the random variable (r.v.) of the number of people in a sample of10 people who have gene A.

X ∼ B(10, 0.24)

From the G.C.,P(X ≤ 4) = 0.933 (3 s.f.)

(i) Let A be the r.v. of the number of people in a sample of 1000people who have gene A.

A ∼ B(1000, 0.24)

Since n = 1000 is large, np = 240 > 5 and npq = 182.4 < 5,

A ∼ N(240, 182.4) approximately

P(230 ≤ A ≤ 260)c.c.= P(229.5 ≤ A ≤ 260.5)

= 0.717 (3 s.f.)

(ii) Let B be the r.v. of the number of people in a sample of 1000people who have gene B.

B ∼ Bin(1000, 0.003)

Since n = 1000 is large and np = 3 < 5,

B ∼ Po(3) approximately

P(2 ≤ B < 5) = P(B ≤ 4)− P(B ≤ 1)

= 0.616 (3 s.f.)



Question 7

Unbiased estimate of population mean, µ,

= x̄

=

∑x

n

=4626

150= 30.84.

Unbiased estimate of population variance, σ2,

= s2

=1

n− 1

(∑x2 − (

∑x)2

n

)=

1

149

(147, 691− (4626)2

150

)= 33.7. (3 s.f.)

H0 : µ = 30H1 : µ > 30

Under H0, test statisticX̄ − 30

s√n

∼ N(0, 1).

We shall perform a one-tail Z-test at 5% level of significance.

x̄ = 30.84.From the G.C., p-value = 0.0382 < 0.05.

Hence, there is sufficient evidence to reject H0 at the 5% level of significance.There is sufficient evidence that the mean time for a student to complete theproject exceeds 30 hours.

Since the sample size (150) is large, X̄ follows a normal distribution by theCentral Limit Theorem. Hence, no assumptions are needed.



Question 8

(i) Let C be the random variable of the cost of a randomly chosenchicken.E(C) = 3(2.2) = 6.6Var(C) = 9(0.52) = 2.25

C ∼ N(6.6, 2.25)

P(C > 7) = 0.395 (3 s.f.)

(ii) Let T be the random variable of the cost of a randomly chosenturkey.E(T ) = 5(10.5) = 52.5Var(T ) = 25(2.12) = 110.25

T ∼ N(52.5, 110.25)

P(C > 7 ∩ T > 55) = (0.395)(0.4059) = 0.160 (3 s.f.)

(iii) E(C + T ) = 52.5 + 6.6 = 59.1Var(C + T ) = 110.25 + 2.25 = 112.5

C + T ∼ N(59.1, 112.5)

P(C + T > 62) = 0.392 (3 s.f.)

(iv) If the price of a chicken exceeds $7 and the price of a turkey exceeds$55, their total price exceeds $62.

However, the converse is not true. For example, a total price of$65 > $62 may be made of up the price of the chicken being $5and the price of the turkey $60.

Hence, the scenario described in part (ii) is a proper subset of thescenario is part (iii). The probability to part (iii) is thus greater.



Question 9

(i) (a) Number of different possible orders = 12! = 479, 001, 600.

(b) We group each couple together and rearrange the 6 groups.Within each group, the man and woman can alternate.

Hence, number of different possible orders= 6!× (2)6 = 46, 080.

(ii) (a) Number of different possible arrangements= (12− 1)! = 39, 916, 800.

(b) We first rearrange the men in a circle. Subsequently, werearrange the women in between.

Hence, number of different possible arrangements= (6− 1)!× 6! = 86, 400.

(c) We group the couples and rearrange them in a circle. Sincethe men and women need to alternate, once we alternatethe man and woman in one group, all other groups mustfollow (there are no additional rearrangements correspond-ing to the remaining 5 groups).

Hence, number of different possible arrangements= (6− 1)!× 2 = 240.

Question 10



Question 10 (cont.)

(i) Using the probability tree,

P(all 3 throws successful) =1

8· 1

4· 1

2=

1

64

(ii) P(at least 2 throws successful) = P(SSS) + P(SSF ) + P(SFS) + P(FSS)

=1

64+

1

8· 1

4· 1

2+

1

8· 3

4· 1

4+

7

8· 1

8· 1

4

=21

256

(iii) P(third throw successful | exactly two throws successful)

=P(third throw successful ∩ exactly two throws successful)

P(exactly two throws successful)

=P(SFS) + P(FSS)

21256− 1

64

=13

17

Question 11

From the G.C., regression line of x on t: x = 66.2− 0.260t (3 s.f.)

When t = 300, x = 66.2− 0.260(300) = −11.8.This value is not desirable as it is negative and concentration values shouldalways be non-negative.The linear model is hence unsuitable to model the relationship between xand t. This can also be seen from the scatter plot, where we can observe adecreasing exponential trend rather than a linear trend.

(i) From the G.C., r = −0.994 (3 s.f.)

As r < 0 and |r| ≈ 1, it shows that y and t have a strong andnegative linear relationship. This value of |r| is also closer to onecompared to value for x and t, indicating that this is a better modelthan the linear model.



Question 11 (cont.)

(ii) From the G.C., regression line of y on t: y = 4.62− 0.0123t (3 s.f.)The y on t line is chosen as t is the independent variable in thisscenario.

When x = 15, y = ln 15.ln 15 = 4.62− 0.0123t∴ t = 155 s (3 s.f.)


h2 / a level mathematics november 2007 paper 2 (9740/2) · h2/ "a" level mathematics...

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